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à 2.7èMore Geometric Proçs for Angles
äèPlease prove ê followïg êorems or show that êy are
not universally true by counterexample.
â
èèèèèèèèè If "hypoêsis," ên "conclusion"
èè Proç:èStatementèèèèèèèèèè Reason
èèèèèè Facts from ê hypoêsisèè Given
èèèèèè Statement ç factsèèèèèèDefïition, axiom, or êorem
éS1 In this section we will look at some proçs about angles that
are similar ë ê proçs ï ê last section about segments.èWe will
need one defïition å an axiom.
è
Defïition 2.7.1èLINEAR PAIR:èA lïear pair ç angles is an adjacent
pair ç angles whose outside rays form a straight angle.
Axiom 15:èIf two angles form a lïear pair, ên êy are supplementary.
èèèèèèèèèèèèèèèèèèèèèèèèConsider ê angles
èèèèèèèèèèèèèèèèèèèèèèèèï ê figure.
@fig2701.BMP,75,175,147,74
Theorem:èIf ▒╖ ïtersects ║┤ at P, ên ╬APB å ╬BPC are supplementary
angles.èAlso, ╬APE å ╬EPC are supplementary.èLikewise ╬EPA å ╬APB,
╬BPC å ╬CPE form supplementary angles.
èProç:èStatementèèèèèèèèèèèèèè Reason
èèèèè1. ▒╖ ïtersects ║┤ at Pèèèèèèè1. Given
èèèèè2. ╬APB ╬BPC form a lïear pairèèè 2. Defïition ç lïear
èèèèèèèèèèèèèèèèèèèèèèèèèè pair
èèèèè3. ╬APB å ╬BPC are supplementaryèè3. (15)Lïear pairs are
èèèèèèèèèèèèèèèèèèèèèèèèèè supplementary
èèèèè4. Similarly ê oêr three pairsèè4. Same reasons as above
èèèèèèè ç angles are supplementary
1èèPlease prove ê given statement is true or show that it isèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèènot universally true by counterexample.èèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèTheorem: If ╬1 ╧ ╬3, ên ╬2 ╧ ╬4.
èèèèèèèèèèèèèèèèè(╬1 å ╬2, ╬3 å ╬4 are supplements)
èèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèè A) True by deductive proçèèè
@fig2702.BMP,35,40,147,74èèèB) Not universally true by counterexample
üèTheorem: If ╬1 ╧ ╬3, ên ╬2 ╧ ╬4
èèèè(╬1 å ╬2, ╬3 å ╬4 are supplements)
èèèèèProç: StatementèèèèèèèèèèèReason
èèèèèèèè 1. ╬1 ╧ ╬3èèèèèèèèèè 1. Givenèèèèèè
èèèèèèèè 2. ╬1 å ╬2, ╬3 å ╬4èèèè2. Given
èèèèèèèèèèèare supplementsèèèèèèèèèèèèèèèèèèèèèèèèèèèments
èèèèèèèè 3. m╬1 + m╬2 = 180°èèèèèè3. Defïition ç
èèèèèèèèèèèèèèèèèèèèèèèèèè supplements
èèèèèèèè 4. m╬3 + m╬4 = 180°èèèèèè4. Defïition ç
èèèèèèèèèèèèèèèèèèèèèèèèèè supplements
èèèèèèèè 5. m╬1 + m╬2 = m╬3 + m╬4èèè 5. Transitive axiom
èèèèèèèè 6. m╬1 + m╬2 = m╬1 + m╬4èèè 6. Substitution
èèèèèèèè 7. m╬2 = m╬4èèèèèèèèè 7. Subtraction axiom for
èèèèèèèèèèèèèèèèèèèèèèèèèè equations
Ç A
2èèPlease prove ê given statement is true or show that it isèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèènot universally true by counterexample.èèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèTheorem: If ╬2 å ╬3, are supplementsè
èèèèèèèèèèèèèèèèèç ╬1, ên ╬2 ╧ ╬3.
èèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèè A) True by deductive proçèèè
@fig2703.BMP,35,40,147,74èèèB) Not universally true by counterexample
üèTheorem: If ╬2 å ╬3 are supplements ç ╬1, ên ╬2 ╧ ╬3.èèè
èèèèèProç: Statementèèèèèèèèèèèè Reason
èèèèèèèè 1. ╬2 å ╬1 are supplementsèèè1. Givenèèèèèè
èèèèèèèè 2. ╬3 å ╬1 are supplementsèèè2. Givenèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèments
èèèèèèèè 3. m╬2 + m╬1 = 180°èèèèèèè 3. Defïition ç
èèèèèèèèèèèèèèèèèèèèèèèèèèèèsupplements
èèèèèèèè 4. m╬3 + m╬1 = 180°èèèèèèè 4. Defïition ç
èèèèèèèèèèèèèèèèèèèèèèèèèèèèsupplements
èèèèèèèè 5. m╬2 + m╬1 = m╬3 + m╬1èèèèè5. Transitive axiom
èèèèèèèè 6. m╬2 = m╬3èèèèèèèèèèè6. Subtraction axiom
èèèèèèèèèèèèèèèèèèèèèèèèèèèèfor equationsèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
Ç A
3èèPlease prove ê given statement is true or show that it isèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèènot universally true by counterexample.èèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèTheorem: If ╬1 ╧ ╬3, ên ╬2 ╧ ╬4.è
èèèèèèèèèèèèèèèèè(╬1 å ╬2, ╬3 å ╬4 are complements)
èèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèè A) True by deductive proçèèè
@fig2704.BMP,35,40,147,74èèèB) Not universally true by counterexample
ü Theorem: If ╬1 ╧ ╬3, ên ╬2 ╧ ╬4.
èèèèèèèè(╬1 å ╬2, ╬3 å ╬4 are complements)èè
èèèè Proç: Statementèèèèèèèèè Reason
èèèèèèèè1. ╬1 ╧ ╬3èèèèèèèèè1. Givenèèèèèè
èèèèèèèè2. ╬1 å ╬2, ╬3 å ╬4èè 2. Given
èèèèèèèèèè are complementsèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèments
èèèèèèèè3. m╬1 + m╬2 = 90°èèèèè3. Defïition ç complementsèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèè4. m╬3 + m╬4 = 90°èèèèè4. Defïition ç complementsèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèè5. m╬1 + m╬2 = m╬3 + m╬4èè5. Transitive axiom
èèèèèèèè6. m╬1 + m╬2 = m╬1 + m╬4èè6. Substitution from 1
èèèèèèèè7. m╬2 = m╬4èèèèèèèè7. Subtraction axiom
èèèèèèèèèèèèèèèèèèèèèèèè for equations
èèèèèèèè8. ╬2 ╧ ╬4èèèèèèèèè8. Defïition ç congruentèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
Ç A
4èèPlease prove ê given statement is true or show that it isèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèènot universally true by counterexample.èèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèTheorem: If ╬2 å ╬3 are complementsè
èèèèèèèèèèèèèèèèèç ╬1, ên ╬2 ╧ ╬3.
èèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèè A) True by deductive proçèèè
@fig2705.BMP,35,40,147,74èèèB) Not universally true by counterexample
ü Theorem: If ╬2 å ╬3 are complements ç ╬1, ên ╬2 ╧ ╬3.èèèèèèèè
èèèè Proç: Statementèèèèèèèèè Reason
èèèèèèèè1. ╬2 å ╬1 areèèèèèè1. Given
èèèèèèèèèè complementsèèèèèè
èèèèèèèè2. ╬3 å ╬1 areèèèèèè2. Given
èèèèèèèèèè complementsèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèments
èèèèèèèè3. m╬2 + m╬1 = 90°,èèèè 3. Defïition ç complements
èèèèèèèèèè m╬3 + m╬1 = 90°èèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèè4. m╬2 + m╬1 = m°3 + m╬1èè4. Transitive axiomèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèè5. m╬2 = m╬3èèèèèèèè5. Subtraction axiom for
èèèèèèèèèèèèèèèèèèèèèèèè equations
èèèèèèèè6. ╬2 ╧ ╬3èèèèèèèèè6. Defïition ç congruentèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
Ç A
5èèPlease prove ê given statement is true or show that it isèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèènot universally true by counterexample.èèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèTheorem: If ╬1 å ╬3 are verticalè
èèèèèèèèèèèèèèèèèangles, ên ╬1 ╧ ╬3.
èèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèè A) True by deductive proçèèè
@fig2706.BMP,35,40,147,74èèèB) Not universally true by counterexample
ü Theorem: If ╬1 å ╬3 are vertical angles, ên ╬1 ╧ ╬3.èèèèèèèè
èèèè Proç: StatementèèèèèèèèèReason
èèèèèèèè1. ╬1 å ╬3 areèèèèè 1. Given
èèèèèèèèèè vertical anglesèèèèè
èèèèèèèè2. ╬1 å ╬2 form aèèèè2. Defïition ç lïear pair
èèèèèèèèèè lïear pairèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè ments
èèèèèèèè3. ╬3 å ╬2 form aèèèè3. Defïition ç lïear pair
èèèèèèèèèè lïear pairèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèè4. ╬1 å ╬2 areèèèèè 4. (15)Lïear pairs are
èèèèèèèèèè supplementsèèèèèèèèsupplementsèèèèèèèèèèèèèèè
èèèèèèèè5. ╬2 å ╬3 areèèèèè 5. (15)Lïear pairs are
èèèèèèèèèè supplementsèèèèèèèèsupplements
èèèèèèèè6. ╬1 ╧ ╬3èèèèèèèè 6. Supplements ç ê same
èèèèèèèèèèèèèèèèèèèèèèèèangle are congruentèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
Ç A
6èèPlease prove ê given statement is true or show that it isèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèènot universally true by counterexample.èèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèTheorem: If ╬1 å ╬2 form a lïearè
èèèèèèèèèèèèèèèèèpair å m╬1 = 90°, ên m╬2 = 90°.
èèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèè A) True by deductive proçèèè
@fig2707.BMP,35,40,147,74èèèB) Not universally true by counterexample
ü Theorem: If ╬1 å ╬2 form a lïear pair å m╬1 = 90°,
èèèèèèèèên ╬2 = 90°.è
èèèè Proç: StatementèèèèèèèèèReason
èèèèèèèè1. ╬1 å ╬2 formèèèèè1. Given
èèèèèèèèèè a lïear pairèèèèèè
èèèèèèèè2. m╬1 = 90°èèèèèèè 2. Givenèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèments
èèèèèèèè3. ╬1 å ╬2 areèèèèè 3. (15)Lïear pairs are
èèèèèèèèèè supplementsèèèèèèèèsupplementsèèèèèèèèèèèèèèèèèèèè
èèèèèèèè4. m╬1 + m╬2 = 180°èèèè4. Defïition ç supplementsèèèèèèèèè
èèèèèèèè5. 90° + m╬2 = 180°èèèè5. Substitution from 2
èèèèèèèè6. m╬2 = 90°èèèèèèè 6. Substraction axiom for
èèèèèèèèèèèèèèèèèèèèèèèèequationsèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
Ç A
7èèPlease prove ê given statement is true or show that it isèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèènot universally true by counterexample.èèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèTheorem: If ╬APE is a straight angle,è
èèèèèèèèèèèèèèèèèên m╬APB + m╬BPC + ╬CPE = 180°.
èèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèè A) True by deductive proçèèè
@fig2708.BMP,35,40,147,74èèèB) Not universally true by counterexample
ü Theorem: If ╬APE is a straight angle, ên m╬APB + m╬BPC +
èèèèèèèèè m╬CPEè=è180°.
èèèè Proç: Statementèèèèèèèèè Reason
èèèèèèèè1. ╬APE is a straightèèè 1. Given
èèèèèèèèèè angleèèèè
èèèèèèèè2. m╬APB + m╬BPC = m╬APCèè2. (12)Angle addition axiomèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèments
èèèèèèèè3. ╬APC å ╬CPE areèèèè3. Defïition ç lïear pair
èèèèèèèèèè lïear pairèèèèèèèèèèèèèèè
èèèèèèèè4. ╬APC å ╬CPE areèèèè4. (15)Lïear pairs are
èèèèèèèèèè supplementsèèèèèèèè supplementsèè
èèèèèèèè5. m╬APC + m╬CPE = 180°èè 5. Defïition ç supplements
èèèèèèèè6. m╬APB + m╬BPC + m╬CPEèè6. Substitution from 2
èèèèèèèèèè = 180°èèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
Ç A
8èèPlease prove ê given statement is true or show that it isèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèènot universally true by counterexample.èèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèèèèTheorem: If ╬1 ╧ ╬4, ên ╬2 ╧ ╬3.è
èèèèèèèè
èèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèèèèèèèèè A) True by deductive proçèèè
@fig2709.BMP,35,40,147,74èèèB) Not universally true by counterexample
ü Theorem: If ╬1 ╧ ╬4, ên ╬2 ╧ ╬3.è
èèèè Proç: StatementèèèèèèèèèèèèReason
èèèèèèèè1. ╬1 ╧ ╬4èèèèèèèèèèè 1. Given
èèèèèèèè2. ╬1 å ╬2 form lïear pairèè2. Givenèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè ments
èèèèèèèè3. ╬3 å ╬4 form lïear pairèè3. Givenèèèèèèèèèèèèèèèèèè
èèèèèèèè4. ╬1 å ╬2 are supplementsèè 4. (15)Lïear pair
èèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèè5. ╬3 å ╬4 are supplementsèè 5. (15)Lïear pairèèèèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèè6. m╬1 + m╬2 = 180°èèèèèèè6. Defïition ç sup.èèèèèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèè7. m╬3 + m╬4 = 180°èèèèèèè7. Defïiiën ç sup.èèèèèèèèèèèèèèèèèèèèèèèèèèè
èèèèèèèè8. m╬1 + m╬2 = m╬3 + m╬4èèèè 8. Transitive axiom
èèèèèèèè9. m╬1 = m╬4èèèèèèèèèè 9. Defïition ç con-
èèèèèèèèèèèèèèèèèèèèèèèèèèègruent
èèèèèèè 10. m╬1 + m╬2 = m╬3 + m╬1èèèè10. Substitution
èèèèèèè 11. m╬2 = m╬3èèèèèèèèèè11. Subtraction axiomèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèèè
Ç A